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If f: $\mathbb{R} \rightarrow \mathbb{R}$, $f'$ exists (finite or infinite) for all reals, and $\lim_{x\to\infty} f'(x) = K$.

Prove $\lim_{x\to\infty} \frac{f(x)}{x} = K$

I started trying to prove it for $K = 0$.

By definition of infinity limit: $\lim_{x\to\infty} f'(x) = K \rightarrow$ given $\epsilon > 0$, $\exists x_{0} \in \mathbb{R}$ such that $\forall x > x_{0}$ we have $|f'(x) - 0| < \epsilon$.

By the Mean Value Theorem: for $a,b > x_{0}$, $\exists c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

Since $c > x_{0}$, we have that $|f'(c)| = |\frac{f(b)-f(a)}{b-a}| < \epsilon$.

not sure where to go from here

...

Conclude: $|f(x)/x| < \epsilon \rightarrow \lim_{x\to\infty} f'(x) = 0 = K$.

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    Consider l'Hospital's rule. Note that this is not the standard $\frac{\infty}{\infty}$-type, but l'Hospital just works well. – 2017-02-01

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Let $\epsilon > 0$. Since $\lim\limits_{x\to \infty} f'(x) = K$, there is a positive number $M$ such that $\lvert f'(x) - K\rvert < \frac{\epsilon}{2}$ whenever $x \ge M$. Let $x > M$. By the mean value theorem, there is a point $c\in (M,x)$ such that $f(x) = f(M) + f'(c)(x - M)$. Write $f'(c) = K + u$ where $\lvert u \rvert < \frac{\epsilon}{2}$. Then

$$\frac{f(x)}{x} - K = \frac{f(M)}{x} - \frac{KM}{x} + u\frac{x-M}{x}.$$

Since $\frac{f(M)}{x} - \frac{KM}{x} \to 0$ as $x\to \infty$, there is a positive number $N$ such that for all $x$, $x > N$ implies $$\left\lvert \frac{f(M)}{x} - \frac{KM}{x}\right\rvert < \frac{\epsilon}{2}.$$ If $x > \max\{M, N\}$, then

$$\left\lvert \frac{f(x)}{x} - K\right\rvert \le \left\lvert \frac{f(M)}{x} - \frac{KM}{x}\right\rvert + \lvert u\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Since $\epsilon$ was arbitrary, $\frac{f(x)}{x}\to K$ as $x\to \infty$.

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    I appreciate your answer. It makes a lot more sense now. When you say $|\frac{f(x)}{x} - K| \leq |\frac{f(M)}{x} - \frac{KM}{x}| + |u|$ why isn't the u a part of the absolute value like this: $|\frac{f(M)}{x} - \frac{KM}{x} + u|$. Did you implicitly use the triangle inequality here? – 2017-02-01
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    @antsatsui yes, the triangle inequality was used. – 2017-02-01
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This is a typical good-part/bad-part argument.

The good part is that where we choose a $c > 0$ and find a $x(c)$ such that $|f'(x)-K| < c$ for $x > x(c)$. The bad part is where $x \le x(c)$.

Then $f(x) =\int_0^x f'(t) dt =\int_0^{x(c)} f'(t) dt +\int_{x(c)}^x f'(t) dt $ so that $f(x)-Kx =\int_0^{x(c)} (f'(t)-K) dt +\int_{x(c)}^x (f'(t)-K) dt =I_1(c)+I_2(c, x) $.

$I_1(c)$ just depends on $c$.

$|I_2(c, x)| \le (x-x(c))c $.

Therefore $|f(x)-Kx| =|I_1(c)+I_2(c, x)| \le|I_1(c)|+|I_2(c, x)| \le |I_1(c)|+|(x-x(c))c| $ so, dividing by $x$, $|\dfrac{f(x)}{x}-K| \le \dfrac1{x}|I_1(c)|+|(1-x(c)/x)c| \lt \dfrac1{x}|I_1(c)|+|c| $.

For any $c > 0$, we can choose $x$ large enough so that $\dfrac1{x}|I_1(c)| < c$ (by choosing $x > \dfrac{I_1(c)}{c}$).

Therefore, for any $c > 0$, for $x > \dfrac{I_1(c)}{c}$, $|\dfrac{f(x)}{x}-K| < 2c$.

Therefore $\lim_{x \to \infty} \dfrac{f(x)}{x} = K$.

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    How do you know that $f'$ is integrable? – 2017-02-01
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    If it isn't, my argument doesn't work. That's a good point. – 2017-02-01